13. (i) In an arithmetic series, the first term is a and the common difference is d - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

To find the total savings of James, we can express his savings as an arithmetic sequence. The first term is £10, and the common difference can be identified from the pattern:
The terms are £10, £9.20, £8.40, which gives us:

• The first term, a = 10
• The second term, a + d = 9.20, thus d = 9.20 - 10 = -0.80
• The n-th term can be written as: $T_n = a + (n-1)d = 10 + (n-1)(-0.80)$
  This means the total amount saved can be expressed as:
  $S_n = \frac{n}{2}[T_1 + T_n] = \frac{n}{2}[10 + (10 + (n-1)(-0.80))] = \frac{n}{2}[20 - 0.8(n - 1)]$
  Setting this equal to £64:
  $\frac{n}{2}[20 - 0.8(n - 1)] = 64$
  This simplifies to:
  $n[20 - 0.8(n - 1)] = 128$
  Solving this equation gives us the quadratic form:
  $n^2 - 26n + 160 = 0$

We can solve the quadratic equation $n^2 - 26n + 160 = 0$ using the quadratic formula:
$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substituting a = 1, b = -26, c = 160:
$n = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 160}}{2 \cdot 1}$
This simplifies to:
$n = \frac{26 \pm \sqrt{676 - 640}}{2} = \frac{26 \pm \sqrt{36}}{2}$
Thus:
$n = \frac{26 \pm 6}{2}$
This gives two potential solutions:
$n = 16\text{ or } n = 10$

James can take either 10 or 16 weeks to save enough money. However, since he needs to reach the exact amount of £64, the number of weeks must not be more than necessary. Therefore, the more reasonable answer is:
James takes n = 10 weeks, as it is the minimal number of weeks required to reach his savings goal.